here is my work.
\(\displaystyle 3x\,−\,2y\,+\,9\,=\,0\)
\(\displaystyle \dfrac{−2y}{−2}\,=\, \dfrac{−3x}{−2}\,−\,\dfrac{9}{−2}\)
\(\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}\)
I think you mean the following:
. . . . .\(\displaystyle \,3x\,−\,3x\,−\,2y\,+\,9\,−\, 9\, =\,0\, −\, 3x\, −\, 9\)
. . . . .\(\displaystyle −2y\, =\, −3x\, −\, 9\)
. . . . .\(\displaystyle −2y\, =\, −3x\, +\, (−9)\)
. . . . .\(\displaystyle \dfrac{−2y}{−2}\, =\, \dfrac{−3x}{−2}\, +\, \dfrac{−9}{−2}\)
But why not make things easier for yourself in the first place?
. . . . .\(\displaystyle 3x\, −\, 2y\, +\, 2y\, +\, 9\, =\, 0\, +\, 2y\)
. . . . .\(\displaystyle 3x\, +\, 9\, =\, 2y\)
Then do the division, but by a positive value, so it's harder to lose signs (as you did in the last quoted line above):
. . . . .\(\displaystyle \dfrac{3x}{2}\, +\, \dfrac{9}{2}\, =\, \dfrac{2y}{2}\)
. . . . .\(\displaystyle \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}\, =\, y\)
\(\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}\)
\(\displaystyle y\, =\, \dfrac{2}{3}\, x\, +\, b\)
How on earth did you get the second line above from the first? Why did the slope change from 3/2 to 2/3? Where did the y-intercept go?
My answer isnt the same as the answer in the book.
Well, yes; once the slope of the original line is lost, the negative reciprocal will be wrong, as must then be the line equation. Try starting again at the last good line (provided to you in corrected form below):
. . . . .\(\displaystyle y\, =\, \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}\)
Please show
all of your steps. Thank you!